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Natural response of a RLC underdamped circuit?

  1. Jan 1, 2012 #1
    Not really sure whether this question belongs here or not (if it doesn't, help move?).
    So I was reading my electronic circuits textbook and am at the section of underdamped RLC unforced response, and the book mentions the natural response as
    [itex]v_{n} = e^{-\alpha\cdot t}(A_{1}\cdot e^{j\omega_{d}t}+A_{2}e^{-j\omega_{d}t})[/itex]
    And then after expanding the equation via Euler's formula, the text book writes
    [itex]v_{n} = e^{-\alpha t}((A_{1}+A_{2})cos(\omega_{d}t)+j(A_{1}-A_{2})sin(\omega_{d}t))[/itex]
    The two steps above i understand, however, the book then writes "Because the unknown constants A1 and A2 remain arbitrary, we replace (A1+A2) and j(A1-A2) with new arbitrary (yet unknown) constants B1 and B2. A1 and A2 must be complex conjugates so that B1 and B2 are real numbers, Therefore, [the above equation] becomes
    [itex]v_{n} = e^{-\alpha t}(B_{1}cos(\omega_{d}t)+B_{2}sin(\omega_{d}t))[/itex]"

    This is where I don't understand. Why (or how) can the imaginary portion of the equation be simply substituted for a real number?

    (Book I'm using: Introduction to Electronic Circuits 8th edition (International Student Version), by Richard C. Dorf, James A. Svoboda Page 380)
  2. jcsd
  3. Jan 1, 2012 #2
    There's a missing j on the sin(). B2 can be real but the sun and cos terms must differ by a factor of j one way or another.
  4. Jan 1, 2012 #3


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    The math is often written like that, but you need to remember that physically it means
    [itex]v_{n} = \Re\left[e^{-\alpha t}(A_{1} e^{j\omega_{d}t}+A_{2}e^{-j\omega_{d}t})\right][/itex] where [itex]\Re[/itex] mean "the real part of".

    To be honest, I don't understand that "explanation" either, and I don't know why the book thinks [itex]A_1[/itex] and [itex]A_2[/itex] are complex conjugates. It doesn't matter if they are or not.

    If you multiply out the real and imaginary parts of
    [itex]A_{1} e^{j\omega_{d}t}+A_{2}e^{-j\omega_{d}t}[/itex],
    where [itex]A_1[/itex] and [itex]A_2[/itex] are complex, and using [itex]e^{jz} = \cos z + j \sin z[/itex],
    you will see that it the real part is of the form [itex]B_1 \cos \omega_d t + B_2 \sin \omega_d t[/itex] where [itex]B_1[/itex] and [itex]B_2[/itex] are real constants.

    (I think Antiphon is wrong about the missing j.)
  5. Jan 1, 2012 #4

    jim hardy

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    i think it's a simple algebraic manipulation...
    Dorf's controls book was known (among us struggling undergrads anyway) for brushing by details when i took that course in 1968.... present day reviews of it on 'net say same thing.

    let's see here

    what Dorf says we need is two real numbers B1 and B2
    such that:
    B1 = A1 + A2
    and :
    B2 = j(A1-A2)


    B1 = A1 + A2
    B2/j = A1 - A2

    add the two above eq's and get
    B1 + B2/j = 2A1;
    or A1 = (B1 +B2/j) /2;;;

    subtract them and get
    B1 - B2/j = 2A2
    or A2 = (B1 - B2/j)/2

    now multiply B2/j term in both those eq's by j/j , giving jB2/j^2, and change the resulting j^2's underneath B2 to minus one's which just flips sign of jB2

    now you have
    A1 = (B1 - jB2)/2
    A2 = (B1 + jB2)/2

    observe A1 and A2 are complex conjugates so Dorf is happy
    and i suppose B could be real
    but i'm not smart enough to know if that's the answer.

    could it be that simple? would any old integer work?

    probably that's what Aleph was saying...

    old jim
    Last edited: Jan 1, 2012
  6. Jan 6, 2012 #5
    Thanks for the replies,
    I asked a friend too, and he said it was just simple algebraic manipulation, so jim_hardy would be correct.

    I really did wish that the authors explain these things more clearly.
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